Write the half reactions for the following redox reactions: (a) \(2 \mathrm{Fe}^{3+}{ }_{\text {(aq. })}+2 \mathrm{I}_{(\mathrm{aq})} \longrightarrow 2 \mathrm{Fe}^{2+}{ }_{\text {(aq.) }}+\mathrm{I}_{2(\mathrm{aq} .)}\) (b) \(\mathrm{Zn}_{(\mathrm{s})}+2 \mathrm{H}_{(\text {aq. })}^{+} \longrightarrow \mathrm{Zn}^{2+}\) (aq.) \(+\mathrm{H}_{2(\mathrm{~g})}\) (c) \(\mathrm{Al}_{(\mathrm{s})}+3 \mathrm{Ag}^{+}{ }_{\text {(aq.) }} \longrightarrow \mathrm{Al}_{\text {(aq.) }}^{3+}+3 \mathrm{Ag}_{(\mathrm{s})}\)

Short Answer

Expert verified
Half-reactions for (a): Oxidation: \(\text{2I}^- \longrightarrow \text{I}_2 + 2 \text{e}^-\), Reduction: \(\text{Fe}^{3+} + \text{e}^- \longrightarrow \text{Fe}^{2+}\); (b): Oxidation: \(\text{Zn} \longrightarrow \text{Zn}^{2+} + 2 \text{e}^-\), Reduction: \(\text{2H}^+ + 2 \text{e}^- \longrightarrow \text{H}_2\); (c): Oxidation: \(\text{Al} \longrightarrow \text{Al}^{3+} + 3 \text{e}^-\), Reduction: \(\text{3Ag}^+ + 3 \text{e}^- \longrightarrow 3\text{Ag}\).

Step by step solution

01

Identify Oxidation and Reduction for each reaction

Analyze each redox reaction to determine which species are being oxidized and which are being reduced. Oxidation involves the loss of electrons, while reduction involves the gain of electrons.
02

Write Half-Reactions for Reaction (a)

For the first reaction, write two half-reactions: one for the oxidation of I- to I2 and one for the reduction of Fe3+ to Fe2+.
03

Balance Atoms and Charges for Reaction (a) Half-Reactions

Balance the atoms and charges for each half-reaction by adding appropriate coefficients and electrons.
04

Write Half-Reactions for Reaction (b)

For the second reaction, write two half-reactions: one for the oxidation of Zn to Zn2+ and one for the reduction of H+ to H2.
05

Balance Atoms and Charges for Reaction (b) Half-Reactions

Balance the atoms and charges for each half-reaction by adding appropriate coefficients and electrons.
06

Write Half-Reactions for Reaction (c)

For the third reaction, write two half-reactions: one for the oxidation of Al to Al3+ and one for the reduction of Ag+ to Ag.
07

Balance Atoms and Charges for Reaction (c) Half-Reactions

Balance the atoms and charges for each half-reaction by adding appropriate coefficients and electrons.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation and Reduction
Understanding the fundamental concepts of oxidation and reduction is crucial to grasp the intricacies of redox reactions. Oxidation is the process where an atom, molecule, or ion loses electrons, while reduction is the gain of electrons. It's essential to note that in a redox reaction, one species is oxidized, and another is reduced.

For instance, when iron (Fe) oxidizes, it loses electrons to form Fe3+, while iodide ions (I-) gain electrons during reduction to form iodine (I2). These simultaneous processes are the hallmarks of redox chemistry and are often remembered by the mnemonic 'LEO the lion says GER': Loss of Electrons is Oxidation, Gain of Electrons is Reduction.

In a classroom setting, visual aids such as electron-dot diagrams can be particularly helpful for students to visualize the transfer of electrons. Also, students should be encouraged to remember that the total number of electrons lost in oxidation must equal the number of electrons gained in reduction.
Balancing Redox Equations
The art of balancing redox equations requires a few systematic steps to ensure that both mass and charge are conserved. The method involves separating the overall chemical equation into two half-reactions: one for oxidation and one for reduction. After identifying these half-reactions, the number of atoms of each element should be balanced, followed by the charge. This is often done by adding H2O, H+, and OH- ions where necessary for balancing atoms, and then adding electrons to balance the charge.

For example, in the oxidation half-reaction of Zn to Zn2+, Zn loses two electrons, which must be accounted for in the half-reaction. On the reduction side, two H+ ions gain these two electrons to form H2 gas. Students frequently find this process challenging, so it is advisable to practice with a variety of redox equations and to utilize step-by-step approaches, such as the ion-electron method or the oxidation number method.
Electrochemistry
The field of electrochemistry is where the concepts of oxidation and reduction find their practical manifestation. Electrochemistry deals with the relationship between electrical energy and chemical changes, which includes the study of redox reactions as they occur in electrochemical cells.

In an electrochemical cell, oxidation occurs at the anode and reduction at the cathode. The movement of electrons from the oxidation half-reaction to the reduction half-reaction generates an electric current that can be harnessed for work. For instance, when aluminum (Al) is oxidized, it forms Al3+ ions, releasing three electrons. These free electrons can then move through an external circuit and reduce Ag+ ions at the cathode, resulting in the deposition of silver (Ag).

By mastering electrochemistry, students can understand how batteries operate, how to prevent corrosion, and even the workings of biological electron transfer processes. Incorporating experiments, such as creating simple voltaic cells, can be an engaging way for students to see electrochemistry in action.

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Most popular questions from this chapter

Identify the substance acting as oxidunt or reductant if any in the following: (i) \(\mathrm{AICl}_{3}+3 \mathrm{~K} \longrightarrow \mathrm{Al}+3 \mathrm{KCI}\) (ii) \(\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{~S} \longrightarrow 3 \mathrm{~S}+\mathrm{H}_{2} \mathrm{O}\) (iii) \(\mathrm{BaCl}_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{BaSO}_{4}+2 \mathrm{NaCl}\) (iv) \(3 \mathrm{I}_{2}+6 \mathrm{NaOH} \longrightarrow \mathrm{NalO}_{3}+5 \mathrm{NaI}+3 \mathrm{H}_{2} \mathrm{O}\)

Select the nature or type of redox change in the following reactions: (ii) \(2 \mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}^{0}\) (b) \(\quad \mathrm{Cl}_{2} \longrightarrow \mathrm{ClO}^{-}+\mathrm{Cl}^{-}\) (c) \(2 \mathrm{KClO}_{3} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{KCl}+3 \mathrm{O}_{2}\) (d) \(\quad\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \longrightarrow \mathrm{N}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{3}+4 \mathrm{H}_{2} \mathrm{O}\) (e) \(10 \mathrm{FeSO}_{4}+2 \mathrm{KMnO}_{4}+8 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{MnSO}_{4}+5 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\) \(\mathrm{K}_{2} \mathrm{SO}_{4}+8 \mathrm{H}_{2} \mathrm{O}\) (f) \(5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+2 \mathrm{KMnO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+\) \(10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\)

$$ \begin{aligned} &\text { Point out the oxidation number of } \mathrm{C} \text { in the following : }\\\ &\mathrm{CH}_{4}, \mathrm{C}_{3} \mathrm{H}_{8}, \mathrm{C}_{2} \mathrm{H}_{6}, \mathrm{C}_{4} \mathrm{H}_{10}, \mathrm{CO}, \mathrm{CO}_{2} \text { and } \mathrm{HCO}_{3}^{-}, \mathrm{CO}_{3}^{2-} \end{aligned} $$

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